# How do you test the improper integral #intx^2 dx# from #(-oo, oo)# and evaluate if possible?

It is divergent.

We deal with the indefinite integral as normal: So:

Then as we are dealing with an infinite integration limit we use the limit definition to get:

Which is clearly divergent (and therefore undefined)

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The integral:

is divergent.

So:

so the improper integral is divergent.

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To test the improper integral ( \int_{-\infty}^{\infty} x^2 , dx ), you would split it into two separate integrals:

[ \int_{-\infty}^{\infty} x^2 , dx = \lim_{a \to -\infty} \int_{a}^{c} x^2 , dx + \lim_{b \to \infty} \int_{c}^{b} x^2 , dx ]

Then, you would evaluate each integral separately and take their limits as ( a ) approaches negative infinity and ( b ) approaches positive infinity. This involves integrating ( x^2 ) with respect to ( x ) and then evaluating the limits of integration.

[ \int_{a}^{c} x^2 , dx = \left[ \frac{x^3}{3} \right]*{a}^{c} ]
[ \int*{c}^{b} x^2 , dx = \left[ \frac{x^3}{3} \right]_{c}^{b} ]

After integrating, you would take the limits of the resulting expressions as ( a ) approaches negative infinity and ( b ) approaches positive infinity. If these limits exist, you can then evaluate them to find the value of the improper integral. If the limits do not exist, the integral diverges.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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